Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <map>
#include <set>
#include <queue>

#define REP(i,k,n) for(int i=k;i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1<<30
#define pb push_back
#define mp make_pair

using namespace std;
typedef long long ll;
typedef pair<int,int> P;

struct edge {
	int from,to;
	int cost;

	edge(int t,int c) : to(t),cost(c) {}
	edge(int f,int t,int c) : from(f),to(t),cost(c) {}

	bool operator<(const edge &e) const {
		return cost < e.cost;
	}
};

vector<edge> G[100005];
int d[100005], d2[100005], cnt[100005];

void dijkstra(int s) {
	priority_queue<P, vector<P>, greater<P> > que;
	rep(i, 100005) d[i] = INF;
	
	que.push(mp(0, s));
	d[s] = 0;

	while(que.size()) {
		P p = que.top(); que.pop();
		int cost = p.first;
		int v = p.second;

		if(d[v] < cost) continue;

		rep(i, G[v].size()) {
			edge e = G[v][i];
			if(d[e.to] > d[v] + e.cost) {
				d[e.to] = d[v] + e.cost;
				que.push(mp(d[e.to], e.to));
			}
		}
	}
}

int main() {
	int n, m;
	cin >> n >> m;

	int s, t;
	cin >> s >> t;
	s--; t--;

	rep(i, m) {
		int a, b;
		cin >> a >> b;
		a--; b--;

		G[a].push_back(edge(b, 1));
		G[b].push_back(edge(a, 1));
	}

	dijkstra(s);
	rep(i, n) d2[i] = d[i];

	dijkstra(t);
	rep(i, n) swap(d[i], d2[i]);

	memset(cnt, 0, sizeof(cnt));
	rep(i, n) {
		if(d2[i] == INF) continue;
		cnt[d2[i]]++;
	}

	ll ans = 0;
	rep(i, n) {
		if(d[i] == INF || d2[i] == INF) continue;
		int x = d[t] - 2 - d[i];

		if(x >= 0) {
			ans += cnt[x];
		}
	}

	cout << ans << endl;

	return 0;
}